The unit vector and its modulus are 1 (i, j, k are the direction vectors on the x, y, and z axes respectively)

Positional relationship of vectors

vector product of vectors

Triangle Rule (connect end to end, connect end to end)

parallelogram rule

Commutative law of addition, associative law, associative law of multiplication, distributive law

The result of dividing a nonzero vector by its modulus is a unit vector in the same direction as the original vector.

The first hexagram, the second hexagram, the third hexagram, the fourth hexagram

The coordinates of the point on the vector

midpoint coordinates between two points

The angles α, β, and γ between the vector and the x, y, and z axes respectively, (cosα)² (cosβ)² (cosγ)²=1

The projection of b on a = b·cosθ

Passing the origin: D=0

Parallel to the coordinate axis

plane through coordinate axis

Parallel to the xoy plane

Substitute the point and eliminate the same element → the general plane equation is transformed into an intercept equation

three forms

Rotation of a line (straight line or curve) about an axis

y²=2x, the busbar is parallel to the z-axis → parabolic cylinder

elliptical cylinder

hyperbolic cylinder

Which axis is the bus parallel to and which axis does not appear?

Surface 1

Surface 2

intersection equation

parametric equations of a circle

point

Direction vector: parallel to the original vector

two o'clock

The denominator of the equation is zero, and the numerator of the corresponding term is also understood to be zero.

Let the two-point equation = t, and sort out the x, y, and z axes containing t

Find the intersection point of the line and the plane, substitute it into the plane equation, and find the t value

Intersection of two planes → solution of two plane equations

line line

Line surface

①Plane beam passing through a straight line

② Assuming that the plane 丄 is known, the normal vectors of the two planes are also perpendicular to each other, so the entry and exit can be found

③The intersection of the plane and the known plane is the projection of the straight line

plane beam equation

Use vertical relationships to find vertical surfaces

①The normal vector of the plane is the direction vector of the point, thus writing the point-direction equation of the straight line

② Find the intersection point of the straight line and the plane, using parametric equations

Find the point-direction equation of the straight line and substitute it into the formula